Suppose you have a Donsker's invariance principle type result given by: \begin{equation*} \frac{1}{\sqrt{\tilde m}}\frac{1}{\hat \sigma_m}\left(\sum_{k = p}^{p + int({\tilde mt})} w_k \right ) \to W(t), ~\text{in the weak sense}, \end{equation*} where W(t) is the standard Brownian motion, and $(w_k)$ are independent and zero centered. It seems that the following is true:
\begin{align*} \frac{1}{\sqrt{\tilde m}} \tilde Q_{m} (t) &\xrightarrow{d} W(t+1) - W(1) \quad \text{where}\\ \tilde Q_{m} (t) &= \frac{1}{\hat \sigma_m}\sum_{k = m + 1}^{p + int({\tilde m(1+t)})} w_k = \frac{1}{\hat \sigma_m}\left(\sum_{k = p}^{p + int({\tilde m(1+t)})} w_k - \sum_{k = p}^{m} w_k\right), ~\text{for}~ t \ge 0. \end{align*}
I can justify that each term converges weakly to $W(t+1)$ and $W(1)$ respectively. But how can I justify that these converge weakly to $W(t+1) - W(1)$? Since the convergence in distribution(weak convergence) is not necessarily closed under addition, I am not sure how to justify this.
However, the joint convergence in distribution $$ (X_n,Y_n) \overset{d}{\longrightarrow} (X,Y), n\to\infty, $$ does imply that $g(X_n,Y_n) \overset{d}{\longrightarrow} g(X,Y), n\to\infty$, for any continuous $g$; in particular, that $X_n+Y_n \overset{d}{\longrightarrow} X+Y, n\to\infty$.
Now, coming back to your question: you speak about Donsker invariance rather than the plain central limit theorem, and it states the functional convergence $$ \left(\frac{1}{\sqrt{\tilde m}}\frac{1}{\hat \sigma_m}\sum_{k = p}^{p + int({\tilde mt})} w_k, t\ge 0 \right ) \overset{d}{\longrightarrow} \big(W(t), t\ge 0\big), n\to\infty, $$ in other words, the joint convergence in distribution for all $t\ge 0$. That said, it does imply the convergence you need.